Answer :

Let ΔABC in which AB, BC and CA are its sides.


To Prove: AB – BC < CA or BC – CA < AB or AC – AB < AB


Construction: Take a point D on AC such that AD = AB and Join B to D


Proof: In ΔABD,


3 is an exterior angle.


So, 3 > 1 …(i)


[Exterior angle of a triangle is greater than one of its opposite interior angle]


Since, AB = AD


2 = 1 …(ii)


[Angle opposite to equal sides are equal]


3 > 2 …(iii)


[from (i) and (ii)]


Now, In ΔBDC,


2 is an exterior angle.


So, 2 > 4 …(iv)


[Exterior angle of a triangle is greater than one of its opposite interior angle]


3 > 4 [from (iii) and (iv)]


BC > CD


[greater angle has a longer side opposite to it]


or CD < BC


or AC – AD < BC [, AD + CD = AC ]


or AC – AB < BC [ by construction, AB = AD]


Hence, the difference of the lengths of any two sides of a triangle is less than the length of the third side.


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